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Question-194963

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Question Number 194963 by C2coder last updated on 20/Jul/23
Answered by witcher3 last updated on 21/Jul/23
∫_0 ^(π/2) tan^(−1) (((2asin(x))/(1−a^2 sin^2 (x))))dx..?
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2asin}\left(\mathrm{x}\right)}{\mathrm{1}−\mathrm{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)}\right)\mathrm{dx}..? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by C2coder last updated on 22/Jul/23
i thought the same anf was able to solve it but i think this form is correct just harder and more complex
$${i}\:{thought}\:{the}\:{same}\:{anf}\:{was}\:{able} \\ $$$$\:{to}\:{solve}\:{it}\:{but}\:{i}\:{think}\:{this}\: \\ $$$${form}\:{is}\:{correct}\:{just}\:{harder}\: \\ $$$${and}\:{more}\:{complex} \\ $$
Answered by witcher3 last updated on 21/Jul/23
∫_0 ^(π/2) tan^(−1) (((2asin^2 (x))/(1−a^2 sin^2 (x))))dx=f(a);f(0)=0w f′(a)=∫_0 ^(π/2) ((2sin^2 (x)(1+a^2 sin^2 (x)))/((1−a^2 sin^2 (x))^2 +(2asin^2 (x))^2 )) =(1/2)∫_0 ^(2π) ((sin^2 (x)(1+a^2 sin^2 (x)))/((−a^2 sin^2 (x)+2aisin^2 (x)+1)(−a^2 sin^2 (x)−2iasin^2 (x)+1))) =2∫_0 ^(π/2) ((sin^2 (x)(1+a^2 sin^2 (x)))/((ia+1)^2 sin^2 (x)+cos^2 (x))((1−ia)^2 sin^2 (x)+cos^2 (x))))dx =2∫_0 ^(π/2) ((tg^2 (x)(1+(1+a^2 )tg^2 (x)))/((1+(ia+1)^2 tg^2 (x))(1+(1−ia)^2 tg^2 (x)))) =∫_(−∞) ^∞ ((t^2 (1+(1+a^2 )t^2 ))/((1+t^2 )(1+(1+ia)^2 t^2 )(1+(1−ia)^2 t^2 ))) _(=g) dt =2iπRes(z,g;im(z)>0) a>0 1+(1+ia)^2 t^2 =0⇒t=(i/(1+ia)) 1+t^2 =0⇒t=i 1+(1−ia)^2 t^2 =0⇒t=(i/(1−ia)) 2iπ.(((−(−a^2 ))/(2i(1−(1−ia)^2 )(1−(1+ia)^2 ))) 2iπ(((−(1/((1+ia)^2 ))(1−(1+a^2 ).((1/(1+ia)))^2 ))/((1−((1/(1+ia)))^2 )(2i(1+ia))(1−(((1−ia)^2 )/((1+ia)^2 ))))) 2iπ(((−(1/((1−ia)^2 ))(1−(1+a^2 ).(1/((1−ia)^2 ))))/((1−(1/((1−ia)^2 )))(1−(((1+ia)/(1−ia)))^2 )(2i(1−ia)))) 2iπ((a^2 /(2i(a^2 +2ia)(a^2 −2ia)))),(π/(a^2 +4)),(π/2)tan^(−1) (a) +2iπ((((2a^2 −2ia))/((−a^2 +2ia)(2i−2a)(4ia))) +2iπ((((2a^2 +2ia))/((−4ia)(−a^2 −2ia)(2i+2a))) 2iπ(((2a−2i)/((2i−a)(2i−2a)(4ia)))−(((2a+2i))/(4ia(−a−2i)(2i+2a))) (π/(2a))((((2a−2i)(−a−2i)(2i+2a)−(2a+2i)(2i−a)(2i−2a))/((a^2 +4)(−4−4a^2 ))) (π/(2a))((((−2a^2 −2ia−4)(2a+2i)−(−2a^2 +2ia−4)(2i−2a))/((a^2 +4)(−4−4a^2 ))) (π/(2a))(((−4a^3 +a^2 (−8i)+a(−4)−8i−(4a^3 +a^2 (−8i)+a(4)−8i))/((a^2 +4)(−4−4a^2 ))) =(π/(a^2 +4))+(π/(2a))(((−8a^3 −8a)/((a^2 +4)(−4−4a^2 )))) =((2π)/(a^2 +4)) f(a)=∫_0 ^a f′(t)dt,∣f(0)=0 =∫_0 ^a ((2π)/(t^2 +4))dt=π∫_0 ^a ((d((t/2)))/((1+((t/2))))=πtan^(−1) ((a/2)) ∫_0 ^(π/2) tan^(−1) (((2asin^2 (x))/(1−a^2 sin^2 (x))))dx=πtan^(−1) ((a/2))
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2asin}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{1}−\mathrm{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)}\right)\mathrm{dx}=\mathrm{f}\left(\mathrm{a}\right);\mathrm{f}\left(\mathrm{0}\right)=\mathrm{0w} \\ $$$$\mathrm{f}'\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2sin}^{\mathrm{2}} \left(\mathrm{x}\right)\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\right)}{\left(\mathrm{1}−\mathrm{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\right)^{\mathrm{2}} +\left(\mathrm{2asin}^{\mathrm{2}} \left(\mathrm{x}\right)\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\right)}{\left(−\mathrm{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{2aisin}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{1}\right)\left(−\mathrm{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{2iasin}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{1}\right)} \\ $$$$ \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\right)}{\left.\left(\mathrm{ia}+\mathrm{1}\right)^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)\right)\left(\left(\mathrm{1}−\mathrm{ia}\right)^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)\right)}\mathrm{dx} \\ $$$$ \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{tg}^{\mathrm{2}} \left(\mathrm{x}\right)\left(\mathrm{1}+\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right)\mathrm{tg}^{\mathrm{2}} \left(\mathrm{x}\right)\right)}{\left(\mathrm{1}+\left(\mathrm{ia}+\mathrm{1}\right)^{\mathrm{2}} \mathrm{tg}^{\mathrm{2}} \left(\mathrm{x}\right)\right)\left(\mathrm{1}+\left(\mathrm{1}−\mathrm{ia}\right)^{\mathrm{2}} \mathrm{tg}^{\mathrm{2}} \left(\mathrm{x}\right)\right)} \\ $$$$=\int_{−\infty} ^{\infty} \frac{\mathrm{t}^{\mathrm{2}} \left(\mathrm{1}+\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right)\mathrm{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\left(\mathrm{1}+\mathrm{ia}\right)^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\left(\mathrm{1}−\mathrm{ia}\right)^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} \right)}\:_{=\mathrm{g}} \mathrm{dt} \\ $$$$=\mathrm{2i}\pi\mathrm{Res}\left(\mathrm{z},\mathrm{g};\mathrm{im}\left(\mathrm{z}\right)>\mathrm{0}\right) \\ $$$$\mathrm{a}>\mathrm{0} \\ $$$$\mathrm{1}+\left(\mathrm{1}+\mathrm{ia}\right)^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} =\mathrm{0}\Rightarrow\mathrm{t}=\frac{\mathrm{i}}{\mathrm{1}+\mathrm{ia}} \\ $$$$\mathrm{1}+\mathrm{t}^{\mathrm{2}} =\mathrm{0}\Rightarrow\mathrm{t}=\mathrm{i} \\ $$$$\mathrm{1}+\left(\mathrm{1}−\mathrm{ia}\right)^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} =\mathrm{0}\Rightarrow\mathrm{t}=\frac{\mathrm{i}}{\mathrm{1}−\mathrm{ia}} \\ $$$$\mathrm{2i}\pi.\left(\frac{−\left(−\mathrm{a}^{\mathrm{2}} \right)}{\mathrm{2i}\left(\mathrm{1}−\left(\mathrm{1}−\mathrm{ia}\right)^{\mathrm{2}} \right)\left(\mathrm{1}−\left(\mathrm{1}+\mathrm{ia}\right)^{\mathrm{2}} \right)}\right. \\ $$$$\mathrm{2i}\pi\left(\frac{−\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{ia}\right)^{\mathrm{2}} }\left(\mathrm{1}−\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right).\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{ia}}\right)^{\mathrm{2}} \right)}{\left(\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{ia}}\right)^{\mathrm{2}} \right)\left(\mathrm{2i}\left(\mathrm{1}+\mathrm{ia}\right)\right)\left(\mathrm{1}−\frac{\left(\mathrm{1}−\mathrm{ia}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{ia}\right)^{\mathrm{2}} }\right)}\right. \\ $$$$\mathrm{2i}\pi\left(\frac{−\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{ia}\right)^{\mathrm{2}} }\left(\mathrm{1}−\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right).\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{ia}\right)^{\mathrm{2}} }\right)}{\left(\mathrm{1}−\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{ia}\right)^{\mathrm{2}} }\right)\left(\mathrm{1}−\left(\frac{\mathrm{1}+\mathrm{ia}}{\mathrm{1}−\mathrm{ia}}\right)^{\mathrm{2}} \right)\left(\mathrm{2i}\left(\mathrm{1}−\mathrm{ia}\right)\right)}\right. \\ $$$$\mathrm{2i}\pi\left(\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2i}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{2ia}\right)\left(\mathrm{a}^{\mathrm{2}} −\mathrm{2ia}\right)}\right),\frac{\pi}{\mathrm{a}^{\mathrm{2}} +\mathrm{4}},\frac{\pi}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{a}\right) \\ $$$$+\mathrm{2i}\pi\left(\frac{\left(\mathrm{2a}^{\mathrm{2}} −\mathrm{2ia}\right)}{\left(−\mathrm{a}^{\mathrm{2}} +\mathrm{2ia}\right)\left(\mathrm{2i}−\mathrm{2a}\right)\left(\mathrm{4ia}\right)}\right. \\ $$$$+\mathrm{2i}\pi\left(\frac{\left(\mathrm{2a}^{\mathrm{2}} +\mathrm{2ia}\right)}{\left(−\mathrm{4ia}\right)\left(−\mathrm{a}^{\mathrm{2}} −\mathrm{2ia}\right)\left(\mathrm{2i}+\mathrm{2a}\right)}\right. \\ $$$$\mathrm{2i}\pi\left(\frac{\mathrm{2a}−\mathrm{2i}}{\left(\mathrm{2i}−\mathrm{a}\right)\left(\mathrm{2i}−\mathrm{2a}\right)\left(\mathrm{4ia}\right)}−\frac{\left(\mathrm{2a}+\mathrm{2i}\right)}{\mathrm{4ia}\left(−\mathrm{a}−\mathrm{2i}\right)\left(\mathrm{2i}+\mathrm{2a}\right)}\right. \\ $$$$\frac{\pi}{\mathrm{2a}}\left(\frac{\left(\mathrm{2a}−\mathrm{2i}\right)\left(−\mathrm{a}−\mathrm{2i}\right)\left(\mathrm{2i}+\mathrm{2a}\right)−\left(\mathrm{2a}+\mathrm{2i}\right)\left(\mathrm{2i}−\mathrm{a}\right)\left(\mathrm{2i}−\mathrm{2a}\right)}{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{4}\right)\left(−\mathrm{4}−\mathrm{4a}^{\mathrm{2}} \right)}\right. \\ $$$$\frac{\pi}{\mathrm{2a}}\left(\frac{\left(−\mathrm{2a}^{\mathrm{2}} −\mathrm{2ia}−\mathrm{4}\right)\left(\mathrm{2a}+\mathrm{2i}\right)−\left(−\mathrm{2a}^{\mathrm{2}} +\mathrm{2ia}−\mathrm{4}\right)\left(\mathrm{2i}−\mathrm{2a}\right)}{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{4}\right)\left(−\mathrm{4}−\mathrm{4a}^{\mathrm{2}} \right)}\right. \\ $$$$\frac{\pi}{\mathrm{2a}}\left(\frac{−\mathrm{4a}^{\mathrm{3}} +\mathrm{a}^{\mathrm{2}} \left(−\mathrm{8i}\right)+\mathrm{a}\left(−\mathrm{4}\right)−\mathrm{8i}−\left(\mathrm{4a}^{\mathrm{3}} +\mathrm{a}^{\mathrm{2}} \left(−\mathrm{8i}\right)+\mathrm{a}\left(\mathrm{4}\right)−\mathrm{8i}\right)}{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{4}\right)\left(−\mathrm{4}−\mathrm{4a}^{\mathrm{2}} \right)}\right. \\ $$$$=\frac{\pi}{\mathrm{a}^{\mathrm{2}} +\mathrm{4}}+\frac{\pi}{\mathrm{2a}}\left(\frac{−\mathrm{8a}^{\mathrm{3}} −\mathrm{8a}}{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{4}\right)\left(−\mathrm{4}−\mathrm{4a}^{\mathrm{2}} \right)}\right) \\ $$$$=\frac{\mathrm{2}\pi}{\mathrm{a}^{\mathrm{2}} +\mathrm{4}} \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\mathrm{a}} \mathrm{f}'\left(\mathrm{t}\right)\mathrm{dt},\mid\mathrm{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{a}} \frac{\mathrm{2}\pi}{\mathrm{t}^{\mathrm{2}} +\mathrm{4}}\mathrm{dt}=\pi\int_{\mathrm{0}} ^{\mathrm{a}} \frac{\mathrm{d}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)}{\left(\mathrm{1}+\left(\frac{\mathrm{t}}{\mathrm{2}}\right)\right.}=\pi\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{a}}{\mathrm{2}}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2asin}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{1}−\mathrm{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)}\right)\mathrm{dx}=\pi\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{a}}{\mathrm{2}}\right) \\ $$

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